Why TM10 and TM01 Do Not Exist in Rectangular W.G?
For a rectangular waveguide, show that the \(TM_{01}\) and \(TM_{10}\) modes do not exist. Also determine whether \(TM_{10}\) is equivalent to \(TM_{01}\), or whether \(TM_{ab}=TM_{ba}\).
Solution
For a TM mode in a rectangular waveguide:
$ H_z=0 $
$ E_z\neq0 $
The longitudinal electric field is given by:
$ E_z = E_{mn} \sin\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right) e^{-j\beta z} $
The sine terms arise from the boundary conditions at the conducting walls. Therefore, both \(m\) and \(n\) must be non-zero for \(E_z\) to exist.
1. Check for \(TM_{10}\)
For the \(TM_{10}\) mode:
$ m=1,\qquad n=0 $
Substituting \(n=0\) into the longitudinal electric field:
$ E_z = E_{10} \sin\left(\frac{\pi x}{a}\right) \sin\left(\frac{0\pi y}{b}\right) e^{-j\beta z} $
Since:
$ \sin(0)=0 $
$ \therefore\quad E_z=0 $
However, a TM mode requires:
$ E_z\neq0 $
Therefore:
$ \boxed{ TM_{10}\text{ does not exist} } $
2. Check for \(TM_{01}\)
For the \(TM_{01}\) mode:
$ m=0,\qquad n=1 $
Substituting \(m=0\):
$ E_z = E_{01} \sin\left(\frac{0\pi x}{a}\right) \sin\left(\frac{\pi y}{b}\right) e^{-j\beta z} $
Since:
$ \sin(0)=0 $
$ \therefore\quad E_z=0 $
Again, this contradicts the TM-mode condition:
$ E_z\neq0 $
Therefore:
$ \boxed{ TM_{01}\text{ does not exist} } $
3. Why Both \(TM_{10}\) and \(TM_{01}\) Do Not Exist
The general condition for TM modes in a rectangular waveguide is:
$ m\neq0,\qquad n\neq0 $
Therefore, the lowest possible values are:
$ m=1,\qquad n=1 $
$ \boxed{ TM_{11} } $
Hence, \(TM_{11}\) is the lowest-order TM mode in a rectangular waveguide.
4. Is \(TM_{10}=TM_{01}\)?
No. \(TM_{10}\) and \(TM_{01}\) are not the same mode. More importantly, in a rectangular waveguide neither of them exists.
The mode indices \(m\) and \(n\) have different physical meanings. The index \(m\) represents the field variation along the \(x\)-direction, which is associated with dimension \(a\), while \(n\) represents the field variation along the \(y\)-direction, which is associated with dimension \(b\).
$ m\rightarrow a $
$ n\rightarrow b $
This can also be seen from the cutoff-frequency equation:
$ f_c = \frac{c}{2} \sqrt{ \left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2 } $
For \(TM_{10}\):
$ f_{c,10} = \frac{c}{2} \sqrt{ \left(\frac{1}{a}\right)^2 + \left(\frac{0}{b}\right)^2 } $
$ f_{c,10}=\frac{c}{2a} $
For \(TM_{01}\):
$ f_{c,01} = \frac{c}{2} \sqrt{ \left(\frac{0}{a}\right)^2 + \left(\frac{1}{b}\right)^2 } $
$ f_{c,01}=\frac{c}{2b} $
If \(a\neq b\), these two cutoff frequencies are different. However, in the TM case, both modes have already been ruled out by the boundary conditions.
5. Is \(TM_{ab}=TM_{ba}\)?
In general:
$ \boxed{ TM_{mn}\neq TM_{nm} } $
The two indices should not simply be exchanged because \(m\) and \(n\) correspond to different directions and dimensions of the rectangular waveguide.
For example:
$ TM_{12} $
has one field variation along the \(x\)-direction and two variations along the \(y\)-direction, whereas:
$ TM_{21} $
has two variations along the \(x\)-direction and one variation along the \(y\)-direction.
$ TM_{12}\neq TM_{21} $
unless the physical situation has a special symmetry, such as \(a=b\). Even in that special case, the field distributions are oriented differently with respect to the coordinate axes.
Final Result
For a rectangular waveguide:
$ \boxed{ TM_{10}\text{ does not exist} } $
$ \boxed{ TM_{01}\text{ does not exist} } $
$ \boxed{ m\neq0,\qquad n\neq0 } $
$ \boxed{ TM_{11}\text{ is the lowest-order TM mode} } $
Also, the mode indices cannot generally be interchanged:
$ \boxed{ TM_{mn}\neq TM_{nm} } $
Therefore, \(TM_{10}\) is not equal to \(TM_{01}\). In fact, both are prohibited modes in a rectangular waveguide.